How do you find the average rate of change of #g(x)=1/(x-2)# over #[x, x+h]# ?

1 Answer
Sep 18, 2017

#-1/((x-2)(x-2+h))#

Explanation:

The average rate of change is the slope of the secant between:

#(x, g(x))# and #(x+h, g(x+h))#

The slope #m# of a line between two points #(x_1, y_1)# and #(x_2, y_2)# is given by the formula:

#m = (y_2-y_1)/(x_2-x_1)#

So the slope of the secant is:

#(g(x+h)-g(x))/((x+h)-x) = (g(x+h)-g(x))/h#

So given:

#g(x) = 1/(x-2)#

its average rate of change over #[x. x+h]# is:

#(1/(color(blue)((x+h))-2)-1/(color(blue)(x)-2))/h = ((color(red)(cancel(color(black)(x)))-color(brown)(cancel(color(black)(2))))-(color(red)(cancel(color(black)(x)))+h-color(brown)(cancel(color(black)(2)))))/(h(x-2)(x+h-2))#

#color(white)((1/(((x+h))-2)-1/((x)-2))/h) = (-color(red)(cancel(color(black)(h))))/(color(red)(cancel(color(black)(h)))(x-2)(x+h-2))#

#color(white)((1/(((x+h))-2)-1/((x)-2))/h) = -1/((x-2)(x-2+h))#

In particular, we find:

#lim_(h->0) (g(x+h)-g(x))/h = -1/(x-2)^2#

This is the slope of #g(x)# at #x#